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2 years ago

Which set of statements explains how to plot a point at the location (-2.5,3.75)?

Mathematics

2 answers:

Brums [2.3K]2 years ago

4 0

Answer:

a is right

Step-by-step explanation:

Mekhanik [1.2K]2 years ago

3 0

Answer:b

Step-by-step explanation:

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Can you please solve and explain how

Karolina [17]

First, it would help to simplify each side more:

Left side: 36 + 3(4x - 9) = 36 + 12x - 27 = 12x + 9

Right side: c(2x + 1) + 25 = 2cx + c + 25

Write the simplified equation:

12x + 9 = 2cx + c + 25

Usually when there is no solution, the coefficients of the variable on both sides are the same, so we can make the coefficient if x on the right side into 12:

2cx >>> 12x

Then, c must equal 6 to make this true.

The answer is choice (C).

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2 years ago

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Fynjy0 [20]

answer is 250 which is D Because 250 of 24%=60, so subtract 60 from 250 it will give you 190

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3 years ago

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A company has a policy of retiring company cars; this policy looks at number of miles driven, purpose of trips, style of car and

pav-90 [236]

Answer:

ans=13.59%

Step-by-step explanation:

The 68-95-99.7 rule states that, when X is an observation from a random bell-shaped (normally distributed) value with mean $\mu$ and standard deviation $\sigma$ , we have these following probabilities

$Pr(\mu - \sigma \leq X \leq \mu + \sigma) = 0.6827$

$Pr(\mu - 2\sigma \leq X \leq \mu + 2\sigma) = 0.9545$

$Pr(\mu - 3\sigma \leq X \leq \mu + 3\sigma) = 0.9973$

In our problem, we have that:

The distribution of the number of months in service for the fleet of cars is bell-shaped and has a mean of 53 months and a standard deviation of 11 months

So $\mu = 53, \sigma = 11$

So:

$Pr(53-11 \leq X \leq 53+11) = 0.6827$

$Pr(53 - 22 \leq X \leq 53 + 22) = 0.9545$

$Pr(53 - 33 \leq X \leq 53 + 33) = 0.9973$

-----------

$Pr(42 \leq X \leq 64) = 0.6827$

$Pr(31 \leq X \leq 75) = 0.9545$

$Pr(20 \leq X \leq 86) = 0.9973$

-----

What is the approximate percentage of cars that remain in service between 64 and 75 months?

Between 64 and 75 minutes is between one and two standard deviations above the mean.

We have $Pr(31 \leq X \leq 75) = 0.9545 = 0.9545$ subtracted by $Pr(42 \leq X \leq 64) = 0.6827$ is the percentage of cars that remain in service between one and two standard deviation, both above and below the mean.